Show that a function $f : X \to [−∞, +∞]$ is measurable with respect to $μ^∗$ if and only if $μ^∗(T)≥μ^∗(T∩{x∈X:f(x)≤a})+μ^∗(T∩{x∈X:f(x)≥b})$

Question

The following is an exercise from Bruckner's Real Analysis:

Measurability can be expressed as a separation property. Let $μ^∗$ be an outer measure on a space $X$. Show that a function $f : X \to [−∞, +∞]$ is measurable with respect to $μ^∗$ if and only if $$μ^∗(T)≥μ^∗(T∩{x∈X:f(x)≤a})+μ^∗(T∩{x∈X:f(x)≥b})$$ for all $T ⊂ X$ and all $−∞ <a<b<+∞.$

I have no idea how to approach especially because of conditions like $a<b$, $T$ may not be measurable, etc.

Any idea/hint?

Answer 1

Here's an idea: $f$ is $\mu^*$-measurable iff $\mu^*(T)=\mu^*(T\cap\{f\leq a\})+\mu^*(T/\{f\leq a\}),\forall T \subset X,a \in \mathbb{R}$. ($\Rightarrow$). Assume $f$ is $\mu^*$-measurable, then $$\mu(T)^*=\mu^*(T\cap\{f\leq a\})+\mu^*(T/\{f\leq a\})=\mu^*(T\cap\{f\leq a\})+\mu^*(T\cap\{f\leq a\}^c)$$ we have $\{f\leq a\}^c \supset \{f\geq b\}$ so $\mu^*(T\cap\{f\leq a\}^c)\geq\mu^*(T\cap\{f\geq b\})$ so $$\mu(T)\geq \mu^*(T\cap\{f\leq a\})+\mu^*(T\cap\{f\geq b\})$$ ($\Leftarrow$). Now assume $\mu(T)\geq \mu^*(T\cap\{f\leq a\})+\mu^*(T\cap\{f\geq b\})$. $$T=T\cap(\{f\leq a\}\cup \{a<f<b\} \cup \{f \geq b\})=(T\cap \{f\leq a\})\cup(T \cap \{a < f < b\})\cup(T\cap \{f\geq b\})$$ We have $$\mu^*(T\cap \{f\leq a\})+\mu^*(T\cap \{f\geq b\})\leq \mu^*(T)\leq \mu^*(T\cap \{f\leq a\})+\mu^*(T \cap \{a < f < b\})+\mu^*(T\cap \{f\geq b\})$$ for $b \to a$ we have $\mu(T)=\mu^*(T\cap\{f \leq a\})+\mu^*(T/\{f \leq a\})$ in the limit.